# Math Help - rational expression

1. ## rational expression

getting some trouble
with dis

square root of x^2+6x=x+square root of 2x

2. Originally Posted by math321
getting some trouble
with dis

square root of x^2+6x=x+square root of 2x
square both sides ...

$x^2 + 6x = x^2 + 2x\sqrt{2x} + 2x$

combine like terms ...

$4x - 2x\sqrt{2x} = 0$

factor ...

$2x(2 - \sqrt{2x}) = 0$

set each factor equal to 0 and solve ... check solutions in the original equation.

3. Originally Posted by math321
getting some trouble
with dis

square root of x^2+6x=x+square root of 2x
$\sqrt{x^2+6x}=x+\sqrt{2x}$

$\sqrt{x^2+6x}=\sqrt{\left(x+\sqrt{2x}\right)^2}$

$\sqrt{x^2+6x}=\sqrt{x^2+2x\sqrt{2x}+2x}$

$6x=2x\sqrt{2x}+2x$

$6x-2x=4x=2x\sqrt{2x}$

Do not do the following....

$\frac{4x}{2x}=2=\sqrt{2x}$

$2=\sqrt{4}=\sqrt{2x}$

as you lose a solution.

$4x-2x\sqrt{2x}=0$

$x\left(4-2\sqrt{2x}\right)=0$

4. wondering what will be wrong in solving algebraically to find x which eventually equals 2? as problem is neither a quadratic equation nor an inequality?

5. Originally Posted by math321
getting some trouble
with dis

square root of x^2+6x=x+square root of 2x
Please use brackets to make you equation unambiguous, the above could plausibly mean either:

square_root(x^2+6x)=x+square_root(2x)

or

square_root(x^2)+6x=x+square_root(2x)

I know that the first of these is the most likely, but someone will always fail to make that assumption and interpret it differently.

CB

6. Originally Posted by hotspice
wondering what will be wrong in solving algebraically to find x which eventually equals 2? as problem is neither a quadratic equation nor an inequality?
It finds the x=2 solution, but not the x=0 solution.
Quadratic equations may be solved by factoring, but factoring when possible, finds all solutions and is not confined to quadratics.

$2x=x\sqrt{2x}$

$x\left(2-\sqrt{2x}\right)=0$

zero multiplied by another value is zero, so the factored version shows that x=0 is also a solution.

If we divide both sides by x, we lose that particular solution, though it finds the x=2 answer.
Also remember that we really should not divide by x because we cannot divide by zero anyway,
so when we do that, we are finding only non-zero solutions.

Do check CaptainBlack's post also.